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Chapter 6: Problem 56

Repeat Problem 55 when \(X\) and \(Y\) are independent exponential random variables, each with parameter \(\lambda=1\).

Short Answer

Expert verified
The probability distribution of the sum of two independent exponential random variables \(X\) and \(Y\) with parameter \(\lambda = 1\) is given by the pdf \(f_Z(z) = e^{-z}(z)\) for \(z \geq 0\).

Step by step solution

01

Find the Probability Density Function (pdf)

The pdf of an exponential random variable with parameter \(\lambda\) is given by: \(f(x) = \lambda e^{-\lambda x}\) Since both \(X\) and \(Y\) are exponential random variables with parameter \(\lambda = 1\), their pdfs are the same: \(f_X(x) = e^{-x}\) for \(x \geq 0\) \(f_Y(y) = e^{-y}\) for \(y \geq 0\)
02

Compute Z=X+Y

We want to find the probability distribution of the sum of the two random variables, which is represented as: \(Z = X + Y\)
03

Determine the pdf of Z through convolution

Convolution is a mathematical operation to determine the pdf of the sum of two random variables. Convolution formula for the sum of two random variables \(X\) and \(Y\) is given by: \(f_Z(z) = \int_{-\infty}^{\infty} f_X(x) f_Y(z-x) dx\)
04

Substitute the pdf of X and Y and simplify

Now we can substitute the pdfs of \(X\) and \(Y\) into the convolution formula, and simplify the expression. Note that since \(f_X(x)\) and \(f_Y(y)\) are both non-zero only for \(x \geq 0\) and \(y \geq 0\), both \(x\) and \(z-x\) must also be non-negative in the integration limits. \(f_Z(z) = \int_{0}^{z} e^{-x} e^{-(z-x)} dx\) Now, we can combine exponents and simplify the expression: \(f_Z(z) = \int_{0}^{z} e^{-z} dx = e^{-z} \int_{0}^{z} dx\) After integrating and simplifying, we get: \(f_Z(z) = e^{-z} (z)\) for \(z \geq 0\) Thus, the pdf of Z is given by \(f_Z(z) = e^{-z} (z)\) for \(z \geq 0\).

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Most popular questions from this chapter

Suppose that \(F(x)\) is a cumulative distribution function. Show that (a) \(F^{\prime \prime}(x)\) and (b) \(1-[1-F(x)]^{n}\) are also cumulative distribution functions when \(n\) is a positive integer. HINT: Let \(X_{1}, \ldots, X_{n}\) be independent random variables having the common distribution function \(\vec{F}\). Define random variables \(Y\) and \(Z\) in terms of the \(X\) so that \(P\\{Y \leq x\\}=F^{n}(x)\), and \(P\\{Z \leq x\\}=1-[1-F(x)]^{n}\).

The random variables \(X\) and \(Y\) have joint density function. $$ f(x, y)=12 x y(1-x) \quad 0

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If \(X\) and \(Y\) are independent standard normal random variables, determine the joint density function of $$ U=X \quad V=\frac{X}{Y} $$ Then use your result to show that \(X / Y\) has a Cauchy distribution.

Repeat Problem 55 when \(X\) and \(Y\) are independent exponential random variables, each with parameter \(\lambda=1\).
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